When the positive integer M is divided by the positive integer y, the

Question

When the positive integer M is divided by the positive integer y, the quotient is 11 and the remainder is z. When you divide z by y, the remainder is 9. Which of the following could be the value of M ?

1. 108
2. 119
3. 20
4. 30

A) 1, 2 and 3
b) 1 and 2 only
c) 3 only
d) 1, 2, 3 and 4
e) 2 only

1. 108
2. 119
3. 20
4. 30

A) 1, 2 and 3
b) 1 and 2 only
c) 3 only
d) 1, 2, 3 and 4
e) 2 only

Nikkb · Accepted Answer

Tricky one if we miss the basics. Find below detailed explanation

When the positive integer M is divided by the positive integer y, the quotient is 11 and the remainder is z. => \(M =11y+z\)
=> Here \(z < y\) (That's the case only then reminder can be z)

When you divide z by y, the remainder is 9. => as \(z < y\) and when z is divided by y remainder is 9 => \(z=9\)

So our equation is \(M =11y +9\) for \(z<y\)

Which of the following could be the value of M ?
So lets write all the numbers in form of above equation. If it satisfies the equation than value can be equal to M
So general steps for such kind of questions : Subtract 9 from number and see if its a multiple of 11. if condition satisfies then number can be value of M

1. 108
=> 108 - 9 =99 so as 99 is multiple of 11 this can be our answer. Wrong
Lets write it in equation form : 108 = 11*9 +9 . So if we compare it with M=11y + z => y=z=9 not possible as z<y always
So this value is not M

2. 119
=> 119 - 9 =110 so as 110 is multiple of 11 this can be our answer.
Lets write it in equation form : 119 = 11*10 +9 . So if we compare it with M=11y + z => y=10 and z =9 => z<y satisfies our equation
So this can be value of M

3. 20
=> 20 - 9 =11 so as 11 is multiple of 11 this can be our answer. Wrong
Lets write it in equation form : 20 = 11*1 +9 . So if we compare it with M=11y + z => y=1 and z =9 => z>y. This can't be true as by our equation z<y always
So this value is not M

4. 30
=> 30 -9 =21 This cant be written in multiple of 11.
So this value is not M

A) 1, 2 and 3
b) 1 and 2 only
c) 3 only
d) 1, 2, 3 and 4
e) 2 only

Answer: E

vishwajeet2015 · Answer

Answer is A .1 2 3 :solution...M/y= 11 +z/y...z/y =9...So m = 11y + 9...choose y =1, 9,10

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bkpolymers1617 · Answer

My friend that's wrong. You did the logic right, but missed an important fundamental. Do you want to give it another shot? Let me know and I ll post the explanation if you need. Thanks

vishwajeet2015 · Answer

pretty sue wnt make dat mistake ever again...thnx a ton

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gracie · Answer

if remainder is less than divisor,
then y>9
let y=10
M=10*11=110+z
only option>110 is 119
E

ScottTargetTestPrep · Answer

Solution:
 
Since z is the remainder when M is divided by y, then z must be less than y. So when z is divided by y and the remainder is 9, it must mean z = 9 (because z/y = 0 R 9). Therefore, M = 11y + 9.

If M = 108, then solving 108 = 11y + 9 yields y = 9. However, 108 divided by 9 will not produce a quotient of 11 and a remainder of 9 (in fact, 108/9 = 12, so the quotient is 12 and the remainder is 0). Thus, M cannot equal 108.

If M = 119, then 119 = 11y + 9 and thus, y = 10. Dividing 9 by 10 produces a remainder of 9 (and a quotient of 0). We see that M could equal 119.

If M = 20, then 20 = 11y + 9 and thus, y = 1. However, 20 divided by 1 will not produce a quotient of 11 and a remainder of 9. Thus, M cannot equal 20.

Finally, if M = 30, then 30 = 11y + 9 does not even produce an integer value for y. Thus, M cannot equal 30 either.

Answer: E

samarpan.g28 · Answer

Firstly, M=11y+z, since y is a positive integer, so z is also a positive integer.
Secondly, z=ky+9, where I am considering k is the quotient. Since y and z, and the remainder 9 are positive integers, therefore, k is a positive integer.

Now, M=11y+ky+9=(11+k)y+9.
Additionally, y>z or, y>ky+9 or, y-ky>9 i.e y(1-k)>9. Note that y and k are positive integers.

If I choose option 2. 119, k can be 0 (that is an integer) and y>9 i.e. y can be 10. So, M=11*10+9=119 where z=9 and since y=10 therefore, y>z. This is a match.
I do not think the above conditions will satisfy any other positive integer values. Option (E) is correct.

Dheeraj_079 · Answer

When the positive integer M is divided by the positive integer y, the quotient is 11 and the remainder is z.
So, M= 11y + z
As the remainder is always lesser than the divisor,
z <y
When you divide z by y, the remainder is 9.
z= ky+9
where k is any constant and the y>9.
Lets take value of y= 10.
Then M = 110 + z where z is lesser than y.
OF the given options, only Option 2 is greater than 110.
Hence the answer is Option e.

When the positive integer M is divided by the positive integer y, the

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